Lincoln invested $930 in an account paying an interest rate of 5.6% compounded daily. Assuming no deposits or withdrawals are made, how long would it take, to the nearest tenth of a year, for the value of the account to reach $2,320?

16.3 years

Step-by-step explanation:

(see attached for reference on compounded interest)

recall that the formula for compound interest is

A = P [1 + (r/n) ]^(nt),

where

A = final amount = given as $2,320

P = principal amount = given as $930

r = rate = given as 5.6% = 0.056

n = number of times compounded per year = 365 since compounded daily

t = time taken in years (we are asked to find this)

Simply substitue the known values above into the equation

A = P [1 + (r/n) ]^(nt)

2320 = 930 [1 + (0.056/365) ]^(365t)

2320 = 930 [1.000153425]^(365t)

2320 / 930 = [1.000153425]^(365t)     (taking log of both sides)

log (2320/930) = log {  [1.000153425]^(365t) }

log (2320/930) = 365t log (1.000153425)

365t = log (2320/930) / log(1.000153425)

365t = 5958.6639

t = 5958.6639/365

t = 16.325 years

t = 16.3 years (to nearest tenth of a year)

this linear equation would have just one solution.

step-by-step explanation:

the reason why this would have one solution is that it is an expression with just one variable. since that variable is not raised to a power, it will always have just one solution.